Mathematics, 23.04.2020 01:19 queenflawless31

Suppose babies born in a large hospital have a mean weight of 33663366 grams, and a variance of 244,036244,036. If 118118 babies are sampled at random from the hospital, what is the probability that the mean weight of the sample babies would differ from the population mean by more than 4545 grams? Round your answer to four decimal places.

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0.3222 = 32.22% probability that the mean weight of the sample babies would differ from the population mean by more than 45 grams.

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal probability distribution

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean $\mu$ and standard deviation(which is the square root of the variance) $\sigma$ , the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean $\mu$ and standard deviation $\sigma$ , the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$ .

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

In this problem, we have that:

$\mu = 3366, \sigma = \sqrt{244036} = 494, n = 118, s = \frac{494}{\sqrt{118}} = 45.48$

Probability if differs by more than 45 grams?

Less than 3366-45 = 3321 or more than 3366+45 = 3411. Since the normal distribution is symmetric, these probabilities are equal. So we find one of them, and multiply by them.

Less than 3321.

pvalue of Z when X = 3321. So

$Z = \frac{X - \mu}{\sigma}$

By the Central Limit Theorem

$Z = \frac{X - \mu}{s}$

$Z = \frac{3321 - 3366}{45.48}$

Z = -0.99

Z = -0.99 has a pvalue of 0.1611

2*0.1611 = 0.3222

0.3222 = 32.22% probability that the mean weight of the sample babies would differ from the population mean by more than 45 grams.

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Suppose babies born in a large hospital have a mean weight of 33663366 grams, and a variance of 244,...

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